Pushing a cabinet on a rough floor

  • Thread starter judas_priest
  • Start date
In summary: Does that mean the force applied would decrease if it wants to keep the speed constant?And also, 7 states "cabinet would accelerate across the floor". That means net force is not 0, hence the acceleration. So, I think I understood the concept, but I'm confused about the language of the options.Any time you exert a force on an object, and the net force is nonzero, the object will accelerate. That's Newton's second law. So what's the net force here? 2* F - F?So is F = a?Also, I still have a doubt about 6th and 7th. If the force applied is twice the original one, then the force of
  • #1
judas_priest
174
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Homework Statement


You are pushing a filing cabinet horizontally across a rough floor (μk = 0.37) in a straight line at a constant speed. Which of the following statements about the magnitudes of the forces acting on the filing cabinet are correct?

1) The force that you exert on the filing cabinet will be more than its weight?

2) The force that you exert on the filing cabinet will be equal to the frictional force on the cabinet

3) The force that you exert on the filing cabinet will be equal to its weight

4) The force that you exert on the filing cabinet will be less than its weight

5) The force that you exert on the filing cabinet will be more than the frictional force on the cabinet

6) If you exerted twice the force, the cabinet would slide at a constant speed that is twice the original value

7) If you exerted twice the force, the cabinet would accelerate across the floor




Homework Equations





The Attempt at a Solution



1 I know is false, because cabinet is moving at a constant speed, hence [itex]F = μMg[/itex]

2 is true because [itex]F = μMg[/itex] (Constant speed)

3 is false. Less than it's weight. Because μ is less than 1

4 true. Because μ is less than 1

5 I have a doubt about this one. Will it be true? Because, to keep it at constant speed A force constant force greater than frictional force is required

6 Not sure

7 Not sure.

Please correct me where I'm wrong and give me an idea about 6th and 7th
 
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  • #2
1. Correct, because ##\mu_k < 1##. 2 to 4 are also correct.

5. How can this be true if 2 is also true?

6. Would there be an unbalanced net force on the cabinet in this case? What does F = ma tell you?

7. See 6.
 
  • #3
Curious3141 said:
1. Correct, because ##\mu_k < 1##. 2 to 4 are also correct.

5. How can this be true if 2 is also true?

6. Would there be an unbalanced net force on the cabinet in this case? What does F = ma tell you?

7. See 6.

I'm assuming 6 is true, 7 is false.
Because if it's moving at twice the constant speed, it's obviously not accelerating.
I had the exact same doubt about 5th. But if it wasn't greater than frictional force, and just equal to it, wouldn't it just stay in equilibrium?
 
  • #4
judas_priest said:
I'm assuming 6 is true, 7 is false.

No! Why would you assume that?

Perhaps it's better to start at the start. What do Newton's first two laws of motion state?

Because if it's moving at twice the constant speed, it's obviously not accelerating.
I had the exact same doubt about 5th. But if it wasn't greater than frictional force, and just equal to it, wouldn't it just stay in equilibrium?

So what exactly is equilibrium? Does it necessarily mean a body is at rest? Can a body be in motion and still remain in equilibrium?
 
  • #5
I'm a terrible teacher
 
  • #6
Curious3141 said:
So what exactly is equilibrium? Does it necessarily mean a body is at rest? Can a body be in motion and still remain in equilibrium?

No, it doesn't mean body is at rest, here equilibrium implies to keep the body at a constant speed AGAINST the frictional force. So does that mean a force that would be equal to frictional force be able to do that?

Also, I'm still not clear about 6th and 7th. If twice the force exerted is applied constantly, it'd slide at a constant speed. (according to first law). But it's no where mentioned that it is applied constantly. Does that mean if twice the force was exerted, the applied force would decrease? Hence, making 6th false and 7th true (because it is decelerating)
 
  • #7
judas_priest said:
No, it doesn't mean body is at rest, here equilibrium implies to keep the body at a constant speed AGAINST the frictional force. So does that mean a force that would be equal to frictional force be able to do that?

Constant velocity (which also implies constant speed), so yes, correct. If the force you exert is equal in magnitude and exactly opposite in direction to the frictional force, the body will be in equilibrium. In this case, equilibrium means the cabinet moves in a straight line at constant speed.

Also, I'm still not clear about 6th and 7th. If twice the force exerted is applied constantly, it'd slide at a constant speed. (according to first law). But it's no where mentioned that it is applied constantly. Does that mean if twice the force was exerted, the applied force would decrease? Hence, making 6th false and 7th true (because it is decelerating)

If twice the force were exerted, what would be the *net* force on the object?

The frictional force remains constant at ##\mu_kMg##.

Initially, the force you applied was the same: ##\mu_kMg##. The net force was zero. This meant the cabinet *was* in equilibrium, traveling at constant velocity.

Now, the force you're applying is ##2\mu_kMg##. So what's the net force on the cabinet now?

What does ##F = ma## tell you?

When ##a## is nonzero, can an object move with constant velocity?
 
  • #8
Curious3141 said:
Constant velocity (which also implies constant speed), so yes, correct. If the force you exert is equal in magnitude and exactly opposite in direction to the frictional force, the body will be in equilibrium. In this case, equilibrium means the cabinet moves in a straight line at constant speed.



If twice the force were exerted, what would be the *net* force on the object?

The frictional force remains constant at ##\mu_kMg##.

Initially, the force you applied was the same: ##\mu_kMg##. The net force was zero. This meant the cabinet *was* in equilibrium, traveling at constant velocity.

Now, the force you're applying is ##2\mu_kMg##. So what's the net force on the cabinet now?

What does ##F = ma## tell you?

When ##a## is nonzero, can an object move with constant velocity?

Are you implying

[tex]F - μkMg= Ma [/tex]?

Hence, [tex] 2μkMg - μkMg = Ma [/tex]

Therefore, a = μkg?
 
  • #9
Okay, so are 6 and 7 true? I just worked out again, and I get them as true.
 
  • #10
But again, 6 says " the cabinet would slide at a constant speed that is twice the original value " . Do they mean twice the original value of speed? How do I prove that? I'm clear about 7th.
 
  • #11
judas_priest said:
But again, 6 says " the cabinet would slide at a constant speed that is twice the original value " . Do they mean twice the original value of speed? How do I prove that? I'm clear about 7th.

If you think the 7th is true, then the 6th certainly can't be true. What do you think?
 
  • #12
Yup, you're right. I still can't understand the 6th one though. Rest is very clear.
 
  • #13
judas_priest said:
Yup, you're right. I still can't understand the 6th one though. Rest is very clear.

If 6th isn't true then you don't have to understand it. Do you? The 7th says the cabinet will accelerate, so it can't have a constant velocity.
 
  • #14
judas_priest said:
Are you implying

[tex]F - μkMg= Ma [/tex]?

Hence, [tex] 2μkMg - μkMg = Ma [/tex]

Therefore, a = μkg?

Yes. ##a = \mu_kg##. (BTW, you should learn to do subscripts properly in LaTex, otherwise it looks like k is a constant you're multiplying everything by. Here, the subscript k refers to "kinetic friction").

So if there's a nonzero acceleration, 7 is clearly true. Agreed?

Can 6 be true if 7 is true?
 

Related to Pushing a cabinet on a rough floor

1. What causes a cabinet to be difficult to push on a rough floor?

The friction between the cabinet's feet and the rough surface of the floor causes resistance, making it difficult to move the cabinet.

2. How can I make pushing a cabinet easier on a rough floor?

Applying a lubricant, such as oil or wax, to the bottom of the cabinet's feet can reduce friction and make it easier to push the cabinet on a rough floor.

3. Is there a weight limit for how much I can push on a cabinet on a rough floor?

Yes, there is a weight limit for pushing a cabinet on a rough floor. The weight limit will depend on the strength of the person pushing and the amount of friction between the cabinet and the floor.

4. Will the type of flooring affect how easy it is to push a cabinet?

Yes, the type of flooring can impact how easy or difficult it is to push a cabinet. Smooth surfaces, such as hardwood or tile, will have less friction than rough surfaces, like carpet or concrete.

5. Can pushing a cabinet on a rough floor cause damage to the cabinet or the floor?

Yes, pushing a cabinet on a rough floor can cause damage to both the cabinet and the floor. The increased friction can cause scratches or wear on the cabinet's feet and the floor's surface.

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